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Zinaida [17]
3 years ago
10

How many moles of air must escape from a 10-m  8.0-m  5.0-m room when the temperature is raised from 0c to 20c? assume the p

ressure remains unchanged at one atmosphere while the room is heated?
Chemistry
1 answer:
slega [8]3 years ago
3 0
Data:

p = 1 atm
V = 10 m * 8 m * 5 m = 400 m^3 = 400,000 liter

To = 0 + 273.15K = 273.15K
Tf = 20 + 273.15K = 293.15K

No - Nf =?

2) Formula

pV = NRT => N = pV / (RT)

3) solution

No = pV / (RTo)

Nf = pV / (RTf)

=> No - Nf = [pv / R] [ 1 / To - 1 / Tf ]

=> No - Nf = [1atm*400,000liter / 0.0821 atm*liter/K*mol ] [ 1 / 273.15 - 1 / 293.15]

No - Nf = 1216.9 moles ≈ 1217 moles

Answer: 1217 moles
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Answer:

6.61 × 10∧-29 m³

Explanation:

Given data:

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2 years ago
The alkali metals react with the halogens to form ionic metal halides. what mass of potassium chloride forms when 5.11 l of chlo
ale4655 [162]
<span>ideal gas law: PV = nRT so .....</span><span> V = PV/(RT) </span>
<span>
Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.
</span><span>
We know the molar mass of K (potassium) = 39.0 g/mol </span>
<span>sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles</span>

<span>Find the balanced equation for the reaction : </span><span>2K + Cl2 → 2KCl </span>
<span>Mole ratio of K:Cl = 2:1 </span>

<span>So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol </span>
<span>
This means that K is in excess but Cl completely reacts. </span>

<span> So we know the mole ratio is  Cl:KCl = 1 : 2 
</span>
<span>Number of moles of Cl (completely) reacted = 0.2053 mol which means the n</span><span>umber of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol </span>

<span>Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol </span>
<span>Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g</span>
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